234. 回文链表

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public boolean isPalindrome(ListNode head) {
        ListNode fast = head;
        ListNode slow = head;
        while(fast != null && fast.next != null){
            fast = fast.next.next;
            slow = slow.next;
        }
        // 链表为奇数个节点
        if(fast != null){
            slow = slow.next;
        }
        fast = head;
        slow = reverse(slow);
        while(slow!=null && fast!=null){
            if(slow.val != fast.val)
                return false;
            slow = slow.next;
            fast = fast.next;
        }
        return true;

    }

    public ListNode reverse(ListNode head){
        ListNode pre = null;
        ListNode cur = head;
        ListNode next;
        while(cur!=null){
            next = cur.next;
            cur.next = pre;
            pre = cur;
            cur = next;
        }
        return pre;
    }
}

题解：
快慢指针+链表反转

先用快慢指针，快指针走两步，慢指针走一步，快指针走完之后，慢指针要不指向后一半的第一个节点（偶数个节点），要不指向最中间的一个节点（奇数个节点），再对后半个链表进行反转，对前后链表进行判断，如果有不一样的值，则不是回文链表。